Conducting waveguides and slab optical waveguides.
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PH 317
February 9, 2002 rev 1
The z-direction is parallel to the axis of the waveguide
Energy in the form of electromagnetic waves is confined within the waveguide
Assume waves with z-dependence of exp(ikzz –it)
The x and y dependence is that of superposed plane waves
These combine to form standing waves between the boundaries in x and y
Solutions will be things like sin (kxx) cos(kyy) exp(ikzz –it)
kx, ky will be determined by boundary conditions
kz will be determined from kz = (2/c2 – kx2 – ky2 ) (or v2 instead of c2 in glass)
First write down all 6 components of both curl equations in free space
Then solve for x and y components of E and B in terms of Ez and Bz or their derivatives
Separate two distinct kinds of waves,
o TE in which Ez is zero so E components are transverse to z
o TM in which Bz is zero, so B components are transverse to z
For TE waves, find the form of Bz which will satisfy the boundary conditions
For TM waves, find the form of Ez which will satisfy the boundary conditions
This will fix the values of kx and ky
In a conducting waveguide with perfect conductors for walls, walls at x=0, x=a, y=0, y=b
o Eparallel = 0, Bperpendicular = 0 at the walls
o at x=0, x=a, Ey = 0, Ez = 0, Bx = 0
o at y=0, y=b, Ex=0, Ez = 0, By = 0
In a slab waveguide, waves travel in the x-z plane and boundaries are at x= -a/2, +a/2
For waves to be confined in the slab, noutside <ninside and  >critical
E and B vectors outside the slab must be exponentially decreasing for the wave to be confined
For both types of waveguides, solve separately for even and odd symmetries within TE and TM
modes
For curl E = -B/t we get
- Ex/y + Ey/x = iBz
(1)
Keeping in mind that /z on any component results in ikz, we have, further
Ey/z - Ez/y = iBx
- ikz Ey +Ez/y = iBx
(2)
Ez/x - Ex/z = iBy
- Ez/x + ikz Ex = iBy
(3)
- Bx/y + By/x = -i/c2 Ez
(4)
- ikz By +Bz/y = -i/c2 Ex
(5)
- Bz/x + ikz Bx = -i/c2 Ey
(6)
For curl B =  E/t, we have
Waveguide Notes PH 317 MJM
February 9, 2002
page 2
To solve for Ex in terms of Bz, Ez and their derivatives, we choose (3) and (5). In (3), Ex appears by itself
(no derivatives) and then for By in (3) we will need (5) where By appears without derivatives.
- Ez/x + ikz Ex = iBy
(3)
- ikz By +Bz/y = -i/c2 Ex
(5)
From this comes
Ex = 1/(ikz) Ez/x + /kz ( /kz/c2 Ex + 1/(ikz)Bz/y ),
Ex = F/(ikz) [Ez/x + /kz Bz/y ],
which simplifies to
where F = 1/(1-k2/kz2), with k = /c .
(7)
This expresses Ex in terms of derivatives of Bz and Ez. Because of (5) we can obtain By likewise:
By = 1/(ikz) Bz/y +/(kzc2) ( F/(ikz) [Ez/x + /kz Bz/y ] or
By = F/(ikz)[ +/(kzc2) Ez/x + Bz/y ]
(8)
Using (2) and (6) we obtain
Ey = F/(ikz) [Ez/y -/kz Bz/x ]
(9)
Bx = F/(ikz) [ Bz/x - /(kzc2) Ez/y ]
(10)
In a conducting waveguide at x=0 and x=a we require Ez=0, Ey=0 and Bx = 0
In a TE mode, Ez=0 everywhere. For Ey=0 and for Bx = 0 we need Bz/x = 0. The function to satisfy
this will be Bz = A cos (kxx), provided that kxa = integer  = n . Thus we require
Bz = A cos (nx/a)
(TE mode)
In a conducting waveguide at y=0 and y=b we require Ez=0, Ex=0 and By = 0. With Ez and its
derivatives all zero in the TE mode, we need Bz/y = 0 at y=0 and y=b. This is satisfied by
Bz = B cos (my/b),
So the overall expression for Bz is
Bz = A cos(nx/a) cos(my/b) exp(ikzz –it) .
From this we derive the four transverse components of E and B, according to (7) – (10) .
PH 317 Waveguide Notes Feb 9 2002
page 3
For a TM mode, Bz and all its derivatives are zero, and we still have to satisfy the same boundary
conditions. Now the requirement at x=0 and x=a for Ey=0 translates into Ez = 0 and Ez/y = 0 .
To satisfy Ez = 0 at x=0 and x=a we may use
Ez = A sin(nx/a) ,
where n = 1,2, ...
and for Ez = 0 at y=0 and y=b we may use
Ez = A sin(nx/a) sin(my/b)
But we must also satisfy Ez/y = 0 at x=0 . Ez/y = m/b A sin(nx/a) cos(my/b) and this is indeed
zero at x=0, so setting Ez =0 satisfied both conditions.
Slab optical waveguide. A slab of width a and index n1 is the waveguide, surrounded on both sides by
material of index n2<n1. For modes to propagate in the slab waveguide we need to launch waves at an
angle greater than the critical angle. This causes the fields in n2 to exponentially decay. We will
separately solve for EVEN waveguide modes where the function of x is cos (kxx) and ODD waveguide
modes where the function of x is sin(kxx)
The sketch shows
a TE ray (Ez=0)
in index n1
reflecting from
a boundary, and
showing a decaying
field in n2. We
will want the
parallel component
of E and the parallel
component of H
to be continuous
across the boundary.
x
n2 < n1
t
Ht
z
x=+a/2
n1
Hi

Hr
All E vectors are in
the -y direction
(into the paper)
The waves travel in the x-z plane.
x= -a/2
The net E field in n1 is
Enet, n1,y = -Ei exp(ikxx + ikzz - it) - Er exp(-ikxx +ikzz -it), or
Enet, n1,y = - [A cos(kxx) +B sin (kxx)] exp( ikzz -it).
We will tackle these one at a time: The EVEN TE mode in the slab waveguide and the ODD TE mode
in the slab waveguide. For the TM modes (Hz = 0) we must also work out both even and odd solutions.
EVEN TE mode conditions.
PH 317 Waveguide Notes Feb 9 2002
page 4
For the parallel component of E at the boundary we have
Ey = -A cos (kx a/2) = -Et exp(i ktx a/2) .
(1)
Since  in n1 is greater than the critical angle, cos t is pure imaginary. And ktx = kt cos t, so ktx is
pure imaginary, or ktx = i kt  where  = | cos ’ | . So the fields in n2 are exponential in x: exp(-kt  x).
We also recall that k1 = /v1 = n1 /c and likewise kt = n2 /c.
For matching the H field parallel components, we want Hz. To obtain Hz we appeal to the curl of E.
(curl E)z = - Hz/t = Ey/x - Ex/y = i Hz .
We apply this both inside and outside
In n1: (then evaluated at x=a/2)
+kx A sin(kxa/2) = i Hz
In n2: (then evaluated at x=a/2)
-Et i ktx exp(iktx a/2) = i Htz .
Setting parallel components of H (namely Hz) equal at the boundary gives
+kx A sin(kxa/2) = -Et i ktx exp(iktx a/2) .
Dividing one equation by the other for Ez and Hz we find
kx tan (kx a/2) = | ktx | = kt  .
This is of the form u tan u = q, where u = kx a/2 and q = | ktx | a/2 = kt  a/2 .
Squaring both sides and adding u2 to both sides gives
u2 tan(u)2 + u2 = (u/cos(u))2 = q2+ u2 .
Next we write out k2 on both sides of the boundary
kt2 = ktz2 + ktx2 = n22 2/c2
k2 = kz2 + kx2 = n12 2/c2
and
.
Because kz is the same on both sides (for constant phase along the boundary) and (kt )2 = -ktx2,
n12 2/c2 - n22 2/c2 = kx2 + (kt )2 .
Multiplying both sides by (a/2)2, we get
u2 + q2 = R2,
where R2 = (a/2)2 (n12 2/c2 - n22 2/c2)
PH 317 Waveguide Notes Feb 9 2002
page 5
Now we want to solve
(u/cos(u))2 = q2+ u2 = R2, where R2 depends only on the slab width and indices.
In particular we want solutions of
cos u = u/R .
But because u tan u = q, we only want solutions where tan u is positive, namely the first and third
quadrants (indeed, all odd quadrants).
One can plot simultaneously cos u and +u/R and -u/R, then accept solutions in the odd quadrants.
Homework:
Due Friday February 15, 2002
No late work accepted. Turn in at the start of class.
1. Show that when u/R = 1 we are at the critical angle from n1 to n2.
2. Work out the condition for odd TE modes (analogous to cos u = u/R , odd quadrants).
3. For a slab whose thickness is 10 microns, at an omega of 1 x 1015 rad/s, and n1 = 1.5 and n2 = 1.46,
determine the valid even TE solutions. Determine the launch angle for each of these (theta in n1). {This
is readily done in Maple. Could also be done on a spreadsheet. }